text:
How does the engine produce aerodynamic lift at high angle of attack?
This answer asserts that "at high angle of attack the engine is producing aerodynamic lift". In the context of that answer, we are concerned about the increased torque that raises the nose of the aeroplane with respect to the center of gravity, so we should narrow down our interpretation of the otherwise vague word "lift" as the force component that is perpendicular to the wing or fuselage of an aeroplane. We define the angle of attack as the angle between the fuselage and the incoming airflow direction at infinity (shorthand word for "reasonably far away"). It is very different from the angle of the fuselage makes with the horizontal at high speed while very close to the latter at very low speed. It puzzles me in two ways.

It claims the engine itself produces fuselage-perpendicular force in addition to the thrust parallel to the fuselage and wing, despite the thrust of the engine is directed parallel to the fuselage and thus the wing. How is that perpendicular force produced?
It claims that fuselage-perpendicular force thus produced is dependent on the angle of attack. How does that work?

I would like to see a technical elucidation of this claim or some references supporting or refuting it, best if they come with mathematical derivation.

Edit:
Some answers below observe that when the plane moves, without engine turned on, against the airflow, part of the airflow curves downwards to conclude that a net force perpendicular to the fuselage is exerted. It is not that simple as looking at part of the flow stream, since part of the flow stream goes upwards above the separating point. The net force is the integration of all the pressure on the whole surface, and the pressure on the opposite side of the object presses in the opposite direction. In fact, in a potential flow, the D'Alembert's paradox  shows that the net pressure is exactly zero. Of course, real airflow is not potential flow, but this refutes the overly simple rationale of using downward airflow to derive the fuselage-perpendicular force.
However, when the engine is turned on, especially at high power. The situation may be different so long as the airflow is consistently flowing in the downward direction long after exiting the engine. The momentum of the airflow through the engine may overwhelm other parts of the airflow. But the argument must be more sophisticated than just the downward direction of the airflow.

"Lift is defined to be the external force exerted perpendicularly on the wing or fuselage of the aeroplane." - I've never heard that definition before; where are you getting it from? As far as I know, lift is defined to be the component of the aerodynamic force which is perpendicular to the relative wind (not the fuselage). Of these two possible meanings of the word "lift", which one are you asking about?

@TannerSwett: I realize now there is ambiguity in the word "lift". I have now edited the question. Please review.

Hans, what exactly do you mean with "It is very different from the angle of the fuselage makes with the horizontal at high speed while very close to the latter at very low speed."? Do you perhaps mean to say that the angle of incidence is negligible at high AoA? I would say the most prominent divergence between pitch and AoA is due to climbing or descending, not speed.

Also, why did you define lift as perpendicular to the wing and fuselage (which, by the way, can differ by an angle called the incidence angle) rather than the more commonly accepted force perpendicular to the freestream velocity vector? (I don't mean to be pedantic, but since you seem to want a precisely defined answer, it might help to precisely define the question as well)

@Sanchises: For the sake of simplicity, I am assuming the axis of the cross section of the wing is parallel to the axis of the plane. The free-stream velocity direction is usually not horizontal. When the plane is moving at high speed and ascending or descending, the angle between the fuselage axis and the free-stream (AOA) is different from that between the fuselage and the horizon. Only when the plane is moving horizontally while its axis is not horizontal (nose up), the two angles are the same. I am trying to distinguish AOA from the angle of the fuselage axis with the horizon. That is all.

Anything, even a metal plate creates "lift" when it is under an angle with an air flow. Lift isn't a magical attribute to "wings", it is just a physical consequence of the law of conservation of momentum: when you push a stream of air down, the reactive force is upwards.
the fuselage/engines also act as minor lifting surfaces: when they are under an angle they, inefficiently, redirect the airflow downwards.
For the engine this is also a bit due to the definition of thrust: thrust equations consider an inlet in the same direction of the thrust vector. So the inlet & compressor redirect the flow "downwards" under high angles of attack. - This effect has to show up somewhere: as a lift force generated by the engine.

You are confusing the concepts. Please see the Note in the last paragraph of my question.

I'm not, as I said indeed there is a vertical component of thrust when it is under an angle. However the trust equation for an engine "expects" a trust that is parallel to the engine direction. Under high angles of attack this no longer holds. So the compressor has to do extra work to align the airflow. - This aligning is however omitted from the trust equation, yet it does provide a force perpendicular to the engine direction - thus a lift force. (both a lifting drag as well as an upwards component).

@Hans: This answer is correct: You have two components, one from pressures on the outside of the nacelle and one from the vertical component of thrust. Mentioning both in one answer does not mean they are confused.

@PeterKämpf and paul23: Right. I mistook angle of attack for the angle of the fuselage with the horizontal line. It should be the angle between the incoming airflow relative to the aeroplane and the fuselage. The two are close when the aeroplane speed is very low. However, your use of the words "up" and "down" lead me to think you are simply referring to the vertical direction relative to gravity, while I am concerned about the direction relative to the plane itself. It would be great if you could modify your first paragraph. Also, I know the force of airflow on a nonparallel plate.

"When you push a stream of air down, the reactive force is upwards". True but it does not account for the net force on an object in a flow stream, since part of the flow stream goes upwards above the separating point. The net force is the integration of all the pressure on the whole surface, and the pressure on the opposite side of the object presses in the opposite direction. In fact, in a potential flow, the D'Alembert's paradox shows that the net pressure is exactly zero. The fluid dynamic force is more complicated than just glancing at part of the flow stream.

@Hans The great thing about physics is that it works independently of the situation. How the airflow is changed doesn't matter (pressure or deflection). The cause of this is indeed pressure distributions, however this "detail" is of little important to the whole "how does x generate lift" question. -- Even worse: you can't use the standard closed form equations to solve this, those equations use a thin body approximation.
An elongated shape under a positive angle of attack will bend the airflow down.

I'd also like to stress that even when the motor is "off", and air is allowed to flow freely through the engine. No fuel is added to the engine so any force in a direction different than parellel to the flight path (drag) cannot be the cause of thrust.
Since the geometry of the intake & whole nacelle does bend the air down. A force is being applied to the airflow, and the aircraft has a resulting reactive force upwards. This is "lift".

My use of the D'Alembert's paradox (what you refer to as closed form equation) is not to derive the force per se but to point out the insufficiency in looking only locally at the streamline as you do in your statement "when you push a stream of air down, the reactive force is upwards". As I pointed out below DeltaLima's answer, this kind of argument works only for large thrust from the engine with the exhaust air speed very high. Suppose we shut off the engine. How do you argue for the extra fuselage-perpendicular force?

Look at the streamline image in https://physics.stackexchange.com/questions/111659/would-an-airplane-flying-through-superfluid-helium-experience-lift-and-drag/126286#126286 labeled Re$\ll$1. The flow is largely downwards with upwards parts. By your argument there should be a net upward force on the airfoil. There is not. It is zero.

I'm having difficulty understanding that question. However I know from personal experiments using a low speed windtunnel that even after stalling there is still positive lift on a wing. - and it can increase again after reaching the "stall minimum" http://users.acsol.net/~nmasters/vortex-lift/lettertoTWITT.html (Given that area is mostly/completely useless for practical purposes).

Thankfully, you have a very clearly worded question. Unfortunately, though, the simple mathematical derivation you desire is not within the scope of a regular answer here (and would be everything but simple).
I agree that some of the other answers confuse lift and thrust, but I think this has more to do with definitions and unclear concepts. So let me first get the definitions out of the way:
Lift is defined as the fraction of the aerodynamic force which is perpendicular to the direction of flow at infinity. This direction is equal (but opposite) to the direction of movement if no wind is present. So per definition the lift of all aircraft components points in the same direction.
Thrust is defined… — well, there is no common definition that would be as simple as the one for lift. Thrust is created by accelerating air. But the same is true for lift, so we need to define a boundary which separates the acceleration which is interpreted as lift and the one which is interpreted as thrust. This is bookkeeping — you need to meticulously add up all pressures acting on all surfaces of the aircraft and then decide which to call lift and which to call thrust.
A lot happens around the intake lips of fan engines, especially at low speed. The intake diameter is dimensioned for a compromise between take-off, when a lot of air is sucked into the engine and high-speed flight when some of the air flowing toward the engine will spill over the intake lips, to flow around the nacelle. In case of the 737 MAX, the compromise was shifted to a smaller diameter in order to keep nacelle size manageable. Therefore, on the 737 MAX you get a lot of suction on those intake lips. The intake is slightly angled in order to support even flow at high angle of attack, so the intake is already doing quite some flow bending. This flow bending is lift - or is it? The vertical component of thrust is also the result of this flow bending.
What is important is where this happens - the engines on the MAX have been shifted quite a bit forward, so this lip suction has a large, destabilizing lever arm. And the suction grows with angle of attack - disproportionally even, due to the angled intake. In the end, this is where most of the destabilizing moment comes from, and it sure is perpendicular to the fuselage.
EDIT:
Now I've had some time for the more mathematical answer desired by @Hans, but I will not dig deeper than fist-order-of-magnitude calculations because solid figures are hard to come by. For example, I did not find the mass flow figures for the 737 MAX's CFM LEAP 1B engines, but this source (thanks @mins!) should be close enough. The 485 kg/s cited there for 145 kN thrust should scale to 436 kg/s for the 130 kN of the LEAP 1B. Next assumption: The airplane's angle of attack is +15° (remember, no flaps when MCAS is active!) so the change in flow direction effected by the intake is 436 kg/s by 15°. This work has to be done at the intake face because once the flow is inside the intake it flows with the angle of the nacelle already. The intake dimensions are the third piece of guesswork here and I assume 1.7 m at the intake face. The last assumption is the air density; I use 1.15 kg/m³ for what follows.
This now allows us to calculate the incoming impulse: 436 kg = 379 m³ flow through an area of π·0.85² = 2.27 m². This needs a flow speed of 167 m/s (Mach 0.5 at sea level - looks very credible).
Next, the intake lips need to add a downward acceleration to a vertical speed component of sin(15°)·167 = 43.2 m/s. The reqired force for this is 436 kg/s·43.2m/s = 18.84 kN, and it all has to act on the intake lips. To put that into perspective: A fully loaded 737 MAX 9 has a mass of maybe 80 tons, so it weighs around 785 kN. Since we have two engines, the total force required for flow bending at the intake faces is 4.8% of the aircraft weight. At the increased lever arm of the new engine location this is already causing a noticeable nose-up pitch.

Without asking for a full mathematical derivation which I'm sure is outside the scope of this answer; do you have an idea of the order of magnitude of this lip suction effect compared to the aerodynamic lift from just the nacelle shape (like in Mike Swonsun's answer)? (For comparison, the latter is apparently about 1% of the total lift on a 747)

Looks like an illuminating answer. Let me look into the intake lip spillage before discussing some details with you.

This intake "suction" may create lift at high AOA inside the nacelle from the top due to flow deflection (like the bottom of a wing) and from the bottom (local vacuum/flow separation) like the top of a wing (sides cancel)? What have the created here?  Where should it be mounted?  Thanks for improving understanding of this, ahh, phenomena.

@RobertDiGiovanni: Please consider that the flow speed inside the duct is considerably higher, so the suction forces on the lower lip and inside the intake are much higher than on the upper lip and the outside of the nacelle when the intake flies at positive angle of attack.

Seeing some similarities here to the Lippisch Aerodyne, where the thrust was directed downward to control pitch.  It may be best for the engine to solve its own problem.  Small elevons in the exhaust could also provide pitch and roll control (differential thrust for yaw) giving aircraft 3 axis control even with complete loss of ailerons, elevator, and rudder.  Hope they can turn this positive.

@ Peter Kampf Yes, and actually that would make 4 lifting surfaces, bottom and top outside, and inside surfaces from intake.

Please see the "Edit" portion of my question.

@Hans: If you find the explanation that lift is created by accelerating air downwards too mundane, I'm afraid I cannot help you. The fuel mass flow is too small to add a significant contribution, so there is only air left on which the aircraft can act. Also, in presence of a running engine you cannot assume potential flow. We are adding energy inside the engine!

Regarding the effect of the thrust of the engine, precisely due to being fully aware that the flow is far from a potential flow, I specifically treat the topic differently and using a separate paragraph, saying that "the situation may be different..." I do not know if you understand my contention, which is that without the engine, the direction of part of the streamline is insufficient to deduce the direction of momentum transfer, partly because the streamlines at other parts run the opposite direction. The d'Alembert paradox gives just such an example. Do you disagree?

@Hans: Since I do not know what image of which streamlines you have in mind, I'm afraid I do not understand the second half of your comment. Did you read my edit? I suggest you take a figurative step back and look at the situation from a larger distance. Less focus on individual streamlines and more on first principles and the overall picture.

The reason I look at the streamline is I find your and others' statement "lift is created accelerating air downwards" vague and misleading. The flow I have in mind is the top image here https://physics.stackexchange.com/a/126286/17354. The streamline above and below the airfoil largely flows downward but upwards at some parts in the front and tail. What can you conclude from the local direction of the streamline? Certainly not that the net force vanishes. I largely agree with @DeltaLima's answer. However I pointed out the subtlety and applicability thereof in my comment below it.

@Hans: The streamlines around that airfoil are from viscous flow at a Reynolds number well below unity, whereas airliner Reynolds numbers are in the tens of millions. That is a completely different regime and what you see in that answer bears little resemblance to the 737 MAX. The answer on Physics SE says as much. And please let me know what about our explanation of lift is misleading - what else is there to create lift from???

You misunderstood my logic. You contend that the direction of the streamline is sufficient to deduce the direction of the net force. I am isolating your criterion out and asking you to put it to work to deduce the net force on that cited diagram (even though with small Reynolds number). I do not see why this criterion works only for high Reynolds number if that is what you will protest. After all, is this not supposed to come from conservation of momentum? What does this have to do with Reynolds number?

@Hans: What diagram? What criterion? I think if you ask more clearly, the answer is easy. Regarding streamlines: You still have to observe continuity: Draw a box around whatever interests you and integrate all inflowing and outflowing mass flows. They must exactly balance. Therefore, the inflowing and outflowing speeds around an airliner must be similar. With those speeds and mass flows you only need to integrate the impulse (look how I did it in the answer) and you get lift. What is missing?

An engine produces additional lift when it is under high angles of attack (and high thrust) because it causes the airflow to curve downwards. For an aircraft that flies a horizontal path at high angle of attack, the airflow approaches the engine horizontally and leaves the engine in a downward angle, close to the negative angle of attack.
In order to make the flow turn downwards, the engine must exert a downward force on the airflow. According to the third law of Newton,there is an opposite force on the engine. The component of this force that is perpendicular to the incoming undisturbed airflow is the additional lift force caused by the engine.

One way to analyse the thrust of an engine is to look at the rate of change of momentum of the air flowing through and around the turbine.
Momentum is the product of mass and velocity: $\vec{p} = m \vec {V}$
As you see, momentum it is a vector quantity.
The rate of change of momentum $\dot{\vec{p}}$ of a mass is equal to force on the object.
By analysing the change of momentum of the air flowing around the engine, we can
determine the thrust vector of the engine.
In the image below, the engine's axle is perfectly aligned with the incoming flow of air. I choose an imaginary volume around the engine, such that the static pressure at the boundary is equal to the static pressure far ahead of the engine. Because the boundary pressure = $p_0$ at every point on the boundary, the integral of pressure over the surface of the volume, the resulting net force would be zero.

The top and bottom boundaries are chosen to be along the streamlines.
The left boundary is experiencing a constant influx of air; the flow is uniform across the left boundary.
The right boundary is experience a constant outflux of air; the flow in non uniform due to the difference in flow velocity in the core and the bypass section of the engine.
The influx of mass through the left boundary is equal to the outflux of mass through the right boundary; I neglect the burned fuel here.
The influx of momentum through the left boundary $\dot{\vec{p}}_{in}$ is equal to $\dot{m}\vec{V} = \iint_{l} \rho (\vec{V}\cdot\hat{n}) \vec{V} dA$
The outflux of momentum through the right boundary $\dot{\vec{p}}_{out}$ is equal to $\iint_{r} \rho (\vec{V}\cdot\hat{n}) \vec{V} dA$
The difference between the inflow of momentum (indicated by the blue vector below the drawing) and the outflow of momentum (indicated by the red vector below the drawing) is the force exerted on the volume of air. The thrust (indicated by the black vector) is the reaction force.

When we now introduce an angle of attack, the bounding volume will change shape. Also the influx and outflux momentum will be different. What is most important is that the thrust generated is no longer purely axial; the thrust vector develops a transverse component. This is the additional lift (and a bit of drag) that the engine creates a high angle of attack.

When the engine is mounted well in front of the centre of gravity, a high angle of attack / high thrust situation will cause an upward pitching moment. This is the case in the Boeing 737 MAX where this effect was changing the handling characteristics at high angles of attack. To make sure the handling would be similar to earlier 737 models, Boeing introduced the - now infamous - Maneuvering Characteristics Augmentation System (MCAS).

is the jet engine not running into the air (what "flow" is being deflected)?  There are 2 sources of thrust here, from the fan and from the jet.  The air is stationary (no kinetic energy) before it is ingested.  But we could mention torque effect around the CG too.

But the (large) nacelle does deflect air downwards.

Good explanation, but two things might be added: At low speed this effect is particularly big due to the high bypass ratio of the new engines, and then you get even more lift from the flow distortion around the intake lips. As shown in your drawing the air passes straight through the intake, but in reality the engine sucks in air from a wider area at low speed, creating additional lift on the intake lips. Oh, and please correct your last sentence: This additional lift is exactly the tilted thrust force vector.

@PeterKämpf you make an interesting point about the lift on the lips. At lower speed this force will increase, but we will see an opposite force on the upper lip. I need to do some additional drawings and calculations tonight (I will include the math in my update). For now I can't (yet) agree on that the additional lift is exactly the tilted thrust for vector.

You say "An engine produces additional lift when it is under high angles of attack (and high thrust) because it causes the airflow to curve downwards." It is not that simple as looking at part of the flow stream, since part of the flow stream goes upwards above the separating point. The net force is the integration of all the pressure on the whole surface, and the pressure on the opposite side of the object presses in the opposite direction. In fact, in a potential flow, the D'Alembert's paradox https://en.wikipedia.org/wiki/D%27Alembert%27s_paradox shows that the net pressure is exactly zero.

@PeterKämpf It seems that we used slightly different definitions of the thrust vector. In the context of this question I had thought of the thrust as the component of total aerodynamics force on the engine that is aligned with the engine's axle. Any transversal forces due to alpha / beta, can than be further decomposed into lift, drag and side force.

You are right, it's a matter of definition, and the way you do it is very legitimate.

+1. It is clear now. I think there is a subtlety here though. The reason this analysis of looking only locally at the inlet  and outlet speeds without working out the flow surrounding the whole object works is that the airflow out of the engine is hypersonic while the flow along the outer surface of the engine is way lower than the sonic speed. Computing the force on an object, e.g. a hollow cylinder, in a subsonic speed airflow, won't be so easy. That was the point of my last comment.

It is not only the engine thrust producing lift at high angles of attack. The lower surfaces of the engine nacelle ALSO produce aerodynamic lift at high angles of attack.
Even at idle thrust there is additional aerodynamic lift being produced. All engine nacelles do this.
The problem with the B737 MAX is that the engine nacelles are placed farther forward and and have a greater lifting “moment” than previous engines used on the B737.

Correct explanation: The nacelles produce aerodynamic lift, just as the fuselage does. However, with idle engines less lift would be produced. A big portion comes from suction on the intake lips at low speed and high power settings. Lip-o-suction, if you want.

Thanks Peter. I have edited my answer.

It is admirable to seek the higher plane of sophistication to gain greater understanding.
First, air is compressible, so D'Alembert is out.  Probably would not be true if velocity was great enough for cavitation, but I did consider sailing battleships blunt end first to see if more energy could be recovered as flow is pulled back into the trailing edge.  There is also some evidence that air curling up and towards the front of a thin undercambered wing (or a jib sail) gives it almost magical lift properties through recovery of drag energy.
But as far as the engine creating lift force 90 degrees from its thrust (making definition of lift even more specific), we may be talking impossibility without deflection of the airstream by some other physical structure.  Tilting the engine out of the direction of travel does create some interesting ideas, but does the pitching or lifting tendency exist without the cowling around the fan?
It will be interesting to see (beyond software) what they do to solve this.  They might want to
take a look at an RB-57 Canberra.

You do not understand the logic of my argument. The use of  D'Alembert paradox is not to use it to compute (the net pressure) but to refute the argument used in many of the answers here.

Thanks anyways.

Also, for compressible fluids, so long as it is inviscid, irrotational and subsonic, the D'Alembert paradox holds, unlike what you claim. The reason for the net pressure in the airflow is that air is viscous, not its compressibility.

Water is inviscid, but air has viscosity.  Oh, please!!!  I love theory, but am more an applications type.  Again, thanks.

Yes, air has viscosity. Do you agree or disagree? Also I have precisely specified the definition of lift. I do not know what your contention is.

It is that viscosity and mass of air have to be considered, and it's variations in pressure (compressivity) is one of the foundations of lift, along with mass flow deflection.  But drag energy recovery (which is what D'Alembert may have been thinking of) is certainly worth study.  I will continue to read yours and Peters writings with interest.


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If the glideslope signal is higher than it should be, can I continue the approach?
When I'm passing Precision Final Approach Fix, if my glideslope is two dot above, and my altitude is at PFAF altitude, can I continue my approach with this glideslope signal?
Is it legal? I think the glideslope indication is not correct with charted PFAF altitude, so I MUST execute missed approach because I can determine that I'm now receiving false glideslope indication.
What statement can I apply in this situation?

What about flying to Localizer-only minimums until you reach the Localizer MAP and going missed from there?

You should continue the approach.  The altitude listed on the PFAF is for standard atmospheric conditions.
We can adjust the barometric altimeter for non-standard pressure using the Kollsman window.
We cannot adjust the barometric altimeter for warmer or colder than standard conditions.  In cold weather, you will intercept the GS inside the point listed on as the FAF for the localizer.  In warmer weather, you will intercept the GS outside the point listed on the localizer.
Due to this issue, the FAA issued a InFO to address pilots who go below MEAs on warmer than standard days.
https://www.faa.gov/other_visit/aviation_industry/airline_operators/airline_safety/info/all_infos/media/2011/InFO11009.pdf

Thx for your answer, too, however this question was not for that point, and I wrote about this above.

If you are at 2,100 FT MSL on a cold day (below -20° C), you may find the GS at full scale deflection crossing ZENZA inbound.  Due to the errors in the altimeter, this is a normal indication.  The GS will intercept at 5.3 NM instead of 5.9 NM.  False glideslope indications are much higher than these.  A 6° false slope crosses ZENZA at 3,800 FT and provides "reverse sensing".  A 9° false slope crosses ZENZA at 5,753 FT MSL with "positive sensing".

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A modern replacement for DH.82's Gipsy Major engine
While I deeply appreciate any vintage aircraft that is powered by their original engines, I was wondering if there are any modern replacement engines available for the de Havilland Tiger Moth aircraft with Gipsy Major engines installed that require no or minor modifications of either the aircraft and/or the engine itself?

That's a pretty unusual configuration these days - 4 inline cylinders, with the cylinders below the endine.
https://en.wikipedia.org/wiki/De_Havilland_Gipsy_Major#/media/File:DHGipsyMajorengineDrover.JPG
Certainly nothing like the flat-4 pancake engines used in modern Cessnas and Pipers.

The Avia (now LOM) M132 4 cylinder inverted inline might fit. Some adaptation work is probably required.

Seems it is used in the Fisher R-80 Tiger Moth. I don't know how I didn't think about that!

No there isn't.  The closest thing would be a Fairchild Ranger inverted in-line 6, almost as old and 70 lbs heavier and I don't think it's ever been tried.
The DHC-1 Chipmunk, DeHavilland Canada's first in-house design (and later license built by mother DH in UK), also used the Major, and a lot of them have been modified with opposed Lycomings, but I can't imaging anyone ruining the lines of a T Moth that way.

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What material is used on the wing surface of this airplane?
This is one of the Redbull's 2015 Flugtag contestants; I was wondering what material is used on the wing surface of this airplane?

video here.

seems normal plastic, why do you think is anything particular?

I was wondering if it is a rated material.

personally, I don't see why it would be, but I'll leave this to others that might have more specific info

Could be something along the lines of clear monofilm like what they make sails out of.

Looks like a biaxially stretched polyester terephthalate foil, like DuPont's Mylar.

I second Peter. There have already been a couple of Flugtag winners using carbon fiber and Mylar wings

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text:
Would a pulsejet/turbine hybrid be efficient?
A pulsejet is an engine with a few moving parts (only the chamber shutters I think) in which fuel is injected in the combustion chamber, leading to a explosion that pushes air to the exhaust. After the explosion, the air flows to the chamber again, leading to a new explosion and the cycle continues.

This type of engine is really inefficient for a number of reasons, but to me this kinda resembles a piston engine; in the sense that it also works by small explosions. Unlike the pulsejet, the piston engines compresses air (the second stage of the combustion cycle on the picture).

From what I could find, in normal turbine engines (like turbofan, turbojet, turboshaft, etc.) the combustion chamber works by a continuous flow of extremely compressed air that continuously expands towards the back, spinning the turbine and continuing the cycle.
Also, I suppose you wouldn't like it to run as fast as a conventional pulsejet (because I've read that the Pulsejets used in the WW2 couldn't last more than a single flight), I would think of small compressor, like a turbocharger, but moved by the turbine (or something like that) in small pulses. But again, I'm only making assumptions.
I tried to search for any kind of research about it, but I could only find this weird turbo-fan concept from a Kiberius Aircraft PDF with the title of "PULSE DETONATION ENGINE TECHNOLOGY: AN OVERVIEW" (the link immediatly downloads the PDF file).

So, in the document says that in the bypass air zone (in the image, page 18) there would be an extra combustion chamber in which would speed up air even more through pulses. But this sounds more like a turbofan-pulsejet modification rather than a pulsejet with a turbine at its end.

It would be nice to include the link to the PDF. Most people are savvy enough these days to note that it's a PDF link and understand that it'll download/open the file on their machine.

Some good thinking here but let's start with the basics:
Pulse jets require a tuned pipe for adequate compression.  The design sends a shockwave back from the exhaust pipe towards the combustion area.  This wave acts as an "invisible piston", acting exactly as a real one in this sense.
Secondly, detonation means blowing something up.  Also means hotter.
If you take a given mixture of fuel and air with excess fuel, it burns slower and cooler.  The flame tends to be orange.  This is known as "rich".
As you increase % of oxygen, the flame becomes blue.  This is known as "lean".  Every welder knows this.
But beyond lean is uncontrollable and dangerous "detonation".  Blows up too fast.  Not good for pistons or turbine blades.
Bottom line is that compression raises efficiency, rendering even the best pulse jets way behind turbines and piston engines.
However, turbines are much lighter and have less maintenance than piston engines.  Turbine efficiency is improving.
Pulse detonation is an exotic attempt to increase the velocity of exhaust gasses for use at hypersonic speeds.  Not exactly needed for your Piper Cub.
Interestingly, tuned pipes can help performance of 2 cycle engines by limiting the loss of fuel/fresh air for the next firing of the cylinder.  They are commonly seen on small R/C aircraft.

Would a pulsejet/turbine hybrid be efficient?

Well, how do you define "efficient"?  Do you have a specific absolute number in mind?  Greater 30% thermal efficiency?  Greater than 50% propulsive efficiency?  It would probably be better to frame this as a relative question, rather than an absolute one.  So then we have to ask, efficient relative to what?
If we ask, "Would a pulsejet/turbine hybrid be more efficient than a pulsejet by itself", well, almost certainly yes.  By adding a bypass fan driven by a turbine, you can move a larger volume of air at a lower exhaust velocity, and thus significantly improve the propulsive efficiency.
However, that would really defeat the point of a pulsejet.  The main advantage of a pulse jet is that it is extremely cheap to both design and manufacture.  You have exactly one moving part (the valve, and maybe a fuel pump too).  Don't have to worry about computational fluid dynamics to design the blade, precision bearings for the rotor, stresses in the rotating parts, high temperature alloys for the turbine blades, etc.  By the time you have constructed a hybrid pulsejet/turbofan, you've introduced all of that mechanical and aerodynamic complexity. At point, the incremental cost to go with a full turbofan is small compared to the huge gain in efficiency that you would get.
Note: the article you linked is about "pulse detonation engine", which are similar to but different than pulsejets.  Pulsejets deflagrate the fuel (subsonic propagation) whereas pulse detonation, as the name implies detonate the fuel (supersonic propagation, resulting in shock waves).


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Why the combustion chambers of a turbine engine aren't used to spin the compressor like a firework wheel?
I searched here in this stack and the only question I found was this one, but I feel like it wasn't very well worded and it is also considered a duplicate. If this question is considered a duplicate of that same question, shouldn't a mod make an edit to that question and write something more akin to this one?

The combustion chamber generates an airflow of combustion strong enough to rotate the turbine blades and eventually the engine itself.
However, if you were to cut out the turbine and attach the combustion chamber to the compressor, would it work the same way?
The closest thing I found to this was this 3D animation of a hypothetical "turbineless jet engine", but no article/research which explored the practicality/efficiency of such an engine.
This is a firework-wheel by the way, the firework rockets are attached in an angle to the axis of rotation:

What do you mean exactly with "cut out the turbine"? You mean compressor + combustion chamber + exhaust instead of compressor + combustion chamber + turbine + exhaust?

@sophit Yes, the combustion chambers would be angled in order to rotate the whole thing.

@sophit The compressor would supply air because the combustion chamber would be attached to it like a turbine, but without the turbine.

@sophit THe combustion chamber, that would also be attached to the main shaft with the compressor, rotating the engine.

The goal of the engine is to power the aircraft forward, not merely sustain itself.  Where would axial thrust come from in your centrifugal combustion example?

@MichaelHall Well, I don't know, thus the question.

Well, there's your answer then.  If you divert all thrust laterally you don't go flying.

@MichaelHall Maybe a turboshaft would benefit from this?

Not to be rude, but highly doubtful.  Reference the last two paragraphs of my answer.

https://en.m.wikipedia.org/wiki/Tip_jet
The link describes the nearest (imho) working application of your idea in aviation.
While it generally works, it goes against important engineering tendencies tthat you need a good reason not to adhere to:

get the complexity out of the rotating parts.
get the mass out of the rotating parts
get the moment of inertia out of the rotating parts
get the asymmetry out of the rotating parts
get the extreme temperatures out of the rotating parts
use fewer moving parts
create as little as possible vortices anywhere

This answer also beautifully illustrates why designers, engineers, and mechanics seem to have such colorful commentary about helicopters.

What you are proposing would certainly be capable of powering a compressor section, but don’t forget that the primary purpose of a jet engine is to propel an aircraft forward by producing thrust along the axis of rotation.
Jet engines happen to generate enough power to tap some energy to run the front compressor section and any fan section, but your suggestion to divert 100% of the combustion energy laterally would seem to be very wasteful.
In addition to all the other engineering problems mentioned in the other answer, there is simply no current problem that this would solve.
Realize that the best engineering minds in the business have been working on engine improvements for decades now, if this was a good idea someone would be using it now.

"your suggestion to divert 100% of the combustion energy laterally would seem to be very wasteful" the combustion chambers the OP is referring to are only slightly turned outward, the biggest part of the thrust would still be axial.

@sophit, Ok, well... I considered the pinwheel example posted in the question and didn't take the time to research what else he might have meant.

Well, nobody else did but apparently the OP has found a satisfying answer anyway

@sophit, well... the question did specify "like a firework wheel". I also asked where axial thrust might come from and the OP didn't have a clue, making no mention of being "only turned slightly outward." So, I just took the question at face value and answered what was asked... Thanks for the insight! ;)

I believe you're suggesting having each combustor attached directly to the turbine shaft and using the hot gas output of each combustor to directly drive the main shaft instead of using the combined gasses flowing through vanes to turn the shaft.
I don't know the math and physics involved, but it just seems that each can individually wouldn't produce as much thrust as they all would together. Maybe I'm totally off on that one.
In your fireworks example, each "combustor" is a small firework that has a self-contained, predetermined amount of "fuel" in it. It burns for a few seconds, then the entertainment is over.
Even if the individual cans, when mounted tangential to the shaft, were to induce the exact same amount of "spin" to the main shaft as is generated by the current setup, the mechanics of getting fuel from the tanks, to a feed rail along the turbine shaft and into a ring spinning at many thousands of RPM to continually feed the combustor cans would be a very difficult feat of engineering. After all, you want the engine to run longer than your fireworks, right?
This suggestion is very impractical from a purely engineering standpoint. While I'm sure the problem could be solved, what would be gained?
Additionally, had the combustor cans been mounted tangentially to the spin of the shaft, the original turbo jet engine would have never worked, since it relied on the exhaust gasses being ejected from the rear of the engine to provide thrust.
A modern, high-bypass ratio fan-jet might be able to work this way, since the vast majority of its thrust is generated by the bypass air going around the jet core. If this were able to spin the fan shaft at sufficient speed (without the fuel lines disintegrating at the required RPM), the fan itself would create a pretty significant amount of thrust, but you'd still lose the thrust of the exhaust gasses themselves.

"what would be gained?"
Well, the turbine needs to be made of superalloys like inconel with ceramic coatings and air vents to cool it off which costs a lot.
Maybe a turbineless jet engine would be a cheaper alternative?

There might be cost savings, however, there would be additional expense in designing, certifying, and maintaining a new system too, @mandiokai. CAAs, airlines and manufacturers are a bit on the risk-adverse side, so unless there are overwhelming benefits, they're a "stay the course" kinda group.

The closest thing I found to this was this 3D animation of a hypothetical "turbineless jet engine", but no article/research which explored the practicality/efficiency of such an engine.

This is a picture of the "turbineless jet engine" as taken from the link in your question:

Even if you could theoretically dispose of the turbine assembly, at the end that design would be more dangerous and less efficient than a standard turbojet. In particular, I see the following limitations (and for sure I'm missing something):

The rotary screw compressor used in the animation is nothing new and is for example normally used to provide compressed air to the jack hammers which are seen on construction sites. It is very good in continually providing a lot of compressed air but as soon as the pressure needs to be high as well (like in a turbojet), then leakages between the spirals and the case would be no more acceptable. Even with a very precise (and costly) manufacturing, the tightness couldn't be ensured due to deformations under load.

Supply fuel to the rotating combustion chambers would need some sort of rotating distribution system. As already said in the other answers, this system has already been tried out in the helicopter world and its tightness has proven to be difficult to achieve and/or unreliable. In case of malfunction, fuel would be dangerously sprayed around.

In a "standard" turbojet, air from the compressor flows straight inside the combustion chamber. Due to the rotation, the air in that design would be highly swirling and getting it inside the combustion chambers in a straight way would be problematic if not impossible. The whole rotating combustion part should also be tight on the left end otherwise the air from the compressor would simply escape from there instead of going through the combustion chambers.

A combustion chamber typically reaches temperatures which can be as high as 1700°C. This value is limited by the thermo-mechanical properties of the material used. This temperature is very important because the higher it is, the more efficient the whole process is. Fancy materials and ways of constructing the blades are continuously being researched exactly to increase that temperature. Making a combustion chamber rotate goes exactly the opposite way: having to resist not only the high temperatures but also the high centrifugal forces, the combustion temperatures should be lowered in order to not overstress the materials. A lower whole efficiency can therefore be expected.

The rotary screw compressor that you linked is a device with *two counterrotating screws", not a single screw with a helical blade like the one in the image. Rotary screw compressors work efficiently, because they trap air in a (moving) chamber that shrinks as it moves towards the exhaust. The helical blade screw, however, relies solely on friction between the outer wall and the pumped medium. As such, it is always highly inefficient, and only really useful when the main goal is to heat the medium, rather than compressing it. And it works abysmally with low viscosity media like air...

@cmaster-reinstatemonica thanks for expanding upon the working principles of the helical screw, which makes it a definitely useless choice for a turbojet

@cmaster-reinstatemonica The helical screw combines heating with compression. That is precisely why it is used in injection molding machines.

@PeterKämpf Exactly what I was thinking about. Fill in cold plastic pellets, get out a warm, homogenized plastic fluid that's exactly at the temperature required for a given viscosity. Ready to fill a mold, or extrude a sheet like in foil/film production.


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Is there a relationship between 2D airfoil aerodynamic results and a revolution body from that airfoil?
The airfoil would be symmetric and the axis of revolution would be its chord line.
Edit1: To clarify the question, suppose we have a NACA 0010. Then, using for example XFoil, I'm able to get, for example, Cd(AoA). Now, I take the airfoil, and I create a body of revolution, with the axis of revolution being the chord line of the airfoil, with the resulting body being like a teardrop. I want to know if there is a way to correlate my 2D Cd(AoA) to the 3D Cd(AoA) of the body I created, so I can predict some of its aerodynamics (with some degree of accuracy) without having to run 3D CFD analysis on it.

Welcome to Av.SE. Is this a homework question, by chance?  What sort of "relationship" are you looking for?

@RalphJ Thank you. It's a question regarding my thesis. I was looking for a relationship for example, from 2D drag coefficient to 3D drag coefficient on the body of revolution

I find the question perfectly clear. At least the thickness of the airfoil will show up in the body of revolution, increasing its aspect ratio and the suction on the surface. Sorry that I cannot give a better answer - a comment leaves preciously little space for a complete answer.

Thank you @PeterKämpf. I'll edit the question so I can clarify it anyway.


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Do left and right mic buttons control both headset microphones?
The aircraft I was flying today has a push to talk integrated button on each control yoke. I was sitting in the right seat, but my headset was plugged into the jacks on the left side of the cockpit. When I pushed the right mic button, the tower could hear a carrier wave but no voice transmission.  It was only when I pushed the left mic button that they could hear my transmission. Is this normal?  Seems like pushing either mic button should allow a transmission from either headset.

Thanks for all these responses. They make perfect sense to me now, although I wish the operation of the G1000 audio functions had been explained to me when I was first getting up to speed on it. They might be buried in the several hundred page operation manual, but I got bogged down in it after the first hundred pages! I'll be sure to let any of my future students know they have to switch plugs when they use a different PTT switch!

This is the expected behaviour. The other pilot might make some sound or even talk to someone else. You don't want any noise interfering with your own voice during transmission.  Two microphones picking up your voice might also decrease the sound quality because of the small delay between the two locations.

The speed of sound over a few feet won't matter at all; you just don't want noise from the other mic. Also, if both pilots try to key up and talk, it's better that one of them (the first) gets heard and the other gets locked out, rahter than ATC or other pilots hearing a garbled mess.

@StephenS: I agree that the extra noise is more a problem than the voice echo, but there is no doubt that 3 to 6 ms multi-path delay (depending on the size of your cockpit) is noticeable and impairs sound quality.

That‘s what is supposed to happen. The person in the right seat might be having a conversation with another passenger and you don’t want their conversation to be transmitted to ATC.
Most audio panels also have a button that lets you isolate the pilot or crew from the rest of the passengers. I use that button all the time when on takeoff and landing so that passengers can continue to talk to each other and I don’t get distracted and miss important checklist items or radio calls.
My audio panel even lets the pilot and co-pilot talk on different frequencies.

We can both be listening and able to talk on Com 1 radio or Com 2 radio. Or the pilot can be on Com 1 and the Co-pilot on Com 2 and vice versa. On my little plane it isn’t too useful, but it can be handy.
The pilot can be monitoring and talking to ATC for flight following while the co-pilot is talking to Flight Services and getting updated weather or filing a PIREP (pilot report of weather). Or if it is a busy frequency, the co-pilot can listen to the ATIS without having to pick out the details over the chatter on the main frequency.
Knowing how this works can be important. I have had the PTT switch fail in flight and the only way to communicate with ATC was to plug the headset into the other set of jacks and use the PTT switch on the other yoke to communicate.

Your audio panel doesn't support a separate mic that either pilot can grab and use in case of  headset failure? My Cessna 177B with PS Engineering intercom (one of the 6000 series) has that.  Also I use two comm's frequently, such as having a pilot passenger grab the ATIS while I monitor ATC when approaching a busy airspace and I don't want to be off frequency.

@CrossRoads My audio panel doesn’t have a mic input, but there is one in the plane. I probably should test it some day to see how it works. I also have a handheld that I can connect to an external-antenna jack in the dash. And I have called them up on my cell phone when the radios failed. So I guess I was overstating it a bit when I said that the only way to communicate was by using the co-pilot set of plugs.

I have a jack for a handheld also, plugging it in connects to one of the antennas and disconnects the panel radio so the handheld isn't fighting the panel radio.


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How does the Angle of attack vary from the root to the tip of a propeller for a fixed pitch prop?
I came across a video while learning about propellers where it showed the angle of attack increased from the root of the propeller to the tip of propeller whose pitch was fixed.
I somehow cannot get my head around it. Is it because the tip moves faster than the root? If so, can anyone explain why moving faster changes the angle of attack at the tip?
I am pretty new to all this so sorry for the stupid question, & I tried searching on the site but could not find answers to this question explicitly.

To be precise, the blade angle is twisted in order to (hopefully) provide a constant angle of attack along the blade. For a fixed pitch prop, this means a single value of forward speed/RPM. For variable pitch, you can control it better.

You are absolutely correct. The "relative wind" at any point on the propeller is a combination of the aircraft's forward motion plus the angular rotation speed of the prop, which is greater at the tips.  This is why, to get the best Angle of Attack, the prop (airfoil) is designed with a twist or "washout" towards the tip.
Notice, with variable pitch props, as the planes forward speed increases, the entire prop AOA
is turned forward to account for change in relative wind.
Fixed pitch props try to find a happy medium between low speed performance and cruise efficiency.  For example, a 10 x 6 (finer pitch) gives better acceleration for take-off, but a 10 x 7 pitch uses less fuel at cruise.

You're thinking of washout,not downwash. Regardless, this refers to a different design consideration (tuning the angle of incidence for benign stall behaviour) rather than a propeller blade which has a twist even for constant angle of incidence.

Edited to washout.  But it is the same thing for a different reason.  A wing washout will change AOA because relative wind is relatively constant across the leading edge (though the other wing can be different in a slip). On a prop washout keeps AOA the same because the angular rotation creates different speeds along the leading edge.  Good catch on the terminology, thanks!

Thanks Robert. So just clarifying here ,that  since the tip Has a higher angular velocity so It's angular rotational velocity vector will have a greater magnitude that at the root.So the resultant(which is the relative wind) of forward velocity vector(which is constant from the root to the tip of the prop) & the rotational velocity vector will Change at the tip & this changes the angle of attack at the tip , right ?

Yes, and I'll add in a consideration of stress load on the prop, which might make them "downwash" it even more, so "pull" or thrust is consistent across prop blade. Wind mill blades topping 60 meters making this an interesting study.

Just clarifying a bit further, the AOA change is actually opposite that stated in the original question about the video.
To quote the explanation on experimentalaircraft.info:
"The design is such that the blade is thick at the hub with a large blade angle and thin at the tip with a low blade angle." or 'washes out' from root to tip, as others have stated.
Thus a blade angle (and resulting AOA) decreasing from the root balances with higher angular rotation speed at the tip to provide a constant (or optimal) thrust along the span of the blade.  Many blades also vary the chord and shape to optimize blade stress and efficiency under local airflow and to counter Mach effects.

Thanks man .The the website also cleared my doubt about why RPM  increases /Engine over revs when the aircraft enters into a dive.


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What is the difference between flutter and vortex induced vibration?
During flutter, the airfoil oscillates with a large amplitude, which generates vortices at the wake. In vortex induced vibration (VIV) the oscillations of the airfoil are caused by the shedding of vortices. Both flutter and VIV have airfoil oscillations and vortex shedding.
How are they different other than the cause and effect relation? And also how to distinguish which one is the cause and which one is the effect for a case where we don't know beforehand whether it is flutter or VIV?

You mean beyond the inverted cause and effect you lined out in your question?

@ AEhere Yes. Is there any other reason than the cause and effect relation? And also how to distinguish which one is the cause and which one is the effect?

Maybe add that bit about distinguishing cause and effect into the question then :) I hope somebody more versed in flutter can help, but from what I recall the answer is along the lines of vortex shedding being what stabilizes the flutter limit cycle oscillations (without the shed vortex the amplitude would diverge). Good question!

@AEhere Thanks for the suggestion. I have updated the question.

I would say that VIV is induced vibrations caused by shed vortices, usually in a control surface,whose energy is totally absorbed and damped by the stiffness of the primary wing or tail structure to which it's connected.  It becomes flutter when the stiffness of the root structure is low so that the vibrating surface is able to displace it in torsion, so that it stores and returns energy, AND the harmonics of the two components align so that they magnify each other. One is just vibrations,"buzz" in other words, and the other induces oscillations in the entire structure that make it come apart.

@JohnK Thanks a lot for the answer. It will be really helpful if you can please elaborate it with a little bit more details.

@DipanjanBarman I've been involved with control surface vibration investigations on airliners in service, working with flutter specialists, so I've picked up some knowledge, but I don't think I'm qualified to post a detailed answer.

Flutter and vortex induced vibrations (VIV) are both dynamic instabilities arising from the interaction of unsteady aerodynamic, inertial and elastic forces. They both extract energy from the air stream to persist.
Instead of perceiving the interaction as aerodynamic forces causing structural motion or structural motion generating aerodynamic forces, it is more meaningful (and realistic) to describe it as a coupling, e.g.,  typically described as coupled differential equations, and observing the differences in how the aerodynamic forces are generated and its components.
For VIV, the wake instability that generates the alternating shed vortices (von Karman vortex street) in turn generates an oscillatory lift coupled with the structural oscillation. So, the wake instability and unsteady effects of flow separation dominate the oscillatory aerodynamic forces.
Flutter does not have to describe a dynamic instability associated with a fluctuating wake, e.g., panel flutter is the dynamic interaction and instability arising out of the out-of-plane motion of an aircraft panel in relation to surface fluctuating pressures. However, stall flutter does arise from flow separation and its resulting unsteady wake.
Wing flutter is the dynamical instability arising from the unsteady aerodynamic forces (including the shape of unsteady wake) coupled with the structural dynamics, however wing flutter does not require massive oscillatory flow separation as with VIV. For attached flows (unstalled), wing flutter could exhibit flutter (at a high enough freestream velocity, i.e., flutter speed) due to the resultant phase difference between the generated aerodynamic forces and structural oscillation creating a positive feedback. The time-varying shape of the wake still plays a role for wing flutter, e.g., the role of the Theodorsen function for unsteady aerodynamic lift generated by structural motion.

Thanks a lot :)


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Are there experimental aircraft regulations in Germany?
The title is the question.
Are there any special regulations regarding experimental aircraft in Germany (or the EU)?
Those "special regulations" would (hopefully) be looser set of regulations in contrast to the LTF-UL regulations for ultralight planes (MTOW < 600kg).

Yes, there are. Everything is regulated in Germany.
Your regular experimental aircraft would have to follow JAR 23 and the LBA has transferred the supervision and certification of experimental aircraft to the Oskar Ursinus Vereinigung (OUV). Before you begin work on your experimental, be sure to contact them and talk to an OUV assessor. This assessor will follow the progress of your work and will regularly check if all work is done properly. You need to prove that you have a heated workshop and document your work sufficiently (like keep samples of every resin mixture that was used for building composite parts). This assessor will also help with contact to experts if you got stuck or with the insane amounts of paperwork, which - this being Germany - often seem baffling and pointless.
You do not need a full certification but at least a load test to 100% design load. The aircraft will get a vorläufige Verkehrszulassung (VVZ; provisional airworthiness certificate) which needs to be renewed regularly. Also, operation of that experimental will be restricted to overflight of sparsely populated areas. But others have succeeded before, so good luck!

Thank you for your answer.
Does if I understand it correctly I could build my own UL and then after having flown it with the help of OUV I can work on getting it set up to abide LTF-UL?

@GittingGud: I'm not so familiar with ULs, but I would expect that JAR23 is replaced by LTF-UL for experimental ULs. Please contact the OUV office (camo(at)ouv.de) for expert advice.

the insane amounts of paperwork, which - this being Germany - often s̶e̶e̶m̶  is baffling and pointless. One of the few things I can't get used to in Germany.

@DeltaLima: I see it this way: Those bureaucrats need something to justify their existence. Better to lock them away in Government offices where they can shuffle paper all day than letting them out into the open.

That makes me smile. Ping me when you're in the Frankfurt (Main) area, I am buying you a beer.


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Why are general aviation planes so expensive?
My question simply is why are GA planes so expensive?

I can understand that an airliner is an incredible money sink simple because of the size, complexity, development cost and especially the certification process.
But I do not completely understand why GA planes are so expensive,
because:

most GA planes are not luxurious
they have considerably less features than modern cars
there shouldn't be a lot of development cost to cover because a lot of GA plane designs are quite old and "simple"

Possible reasons I can see are:

there are no cheap regulations/certifications for privately used
planes
the overall production amount is not enough to profit from the
economy of scale

Further clarification:

I'm interested in the reasons for purchase and operating cost.
To narrow down the scope of the question I mean planes in the class
of a Cessna 172
location of interest is Europe/Germany but a more general answer would be appreciated to give the question/answer more weight

Are you talking about the purchase price, or operating price? Because, GA planes are much cheaper than I would have imagined. A used 172 can be bought for the price of a mid-range car.

@Rob I think both purchase and operating price should have the same reasons. If not the explanation for the difference (with numbers) would make a great answer.

Location and aircraft category would help answer this; GA is a nice catch-all but there is a large difference between an ultralight, an SLSA and a Part 23 craft. For the first two categories, you'd be surprised by what gets to fly.

@AEhere I edited to question to narrow the scope to "Cessna 172'like" planes in Europe. And I will properly ask a question on the lines of "how much money do I need to spend to fly with reasonable safety"(to go into UL and SLSA).

Related: Is there a good source for estimating the per unit cost breakdown for a classic GA plane like the Cessna 172?

Purchase price
Airplanes are times more complex than cars.
The price of an aircraft is not bound solely to its design. Even though not luxurious as an A380, general aviation planes are still expensive because they must also be reliable. You can't just build an aircraft and start selling it. It must be certified by international organizations, such as EASA and FAA, and that is not cheap.

Before a newly developed aircraft model may enter into operation, it must obtain a type certificate from the responsible aviation regulatory authority. Since 2003, EASA is responsible for the certification of aircraft in the EU and for some European non-EU Countries. This certificate testifies that the type of aircraft meets the safety requirements set by the European Union. (source)

Not only that costs money, but also time. On average, a new GA aircraft will take 3 years to be certified (source).
Not just the certification organizations fees that are high, but to test requires multiple prototypes. And if the plane fails any safety assertions, then it goes back to design phase. Every single component, from the flaps to the belts, must be tested under all kinds of conditions.
The demand also counts. Unlike cars, most people don't buy planes because they don't have to. And even if they buy, they don't usually change their planes every year. For example, Cirrus SR22 was first produced on 2001 and until 2016 only ~5000 units were sold (source). It took 15 years to sell 5000 units. That makes this aircraft to cost 629,900 USD (2019).
But still, you can find a Cessna 172 for the price of a 2019 Lamborghini Urus (200,000 USD).
Operating cost
Ok, so you signed the deal and bought your Cessna 172. Now you have to deal with maintenance, fuel, oil, landing fees, hangar rental and all. And of course, insurance!

Maintenance (fuel, oil, landing fees): 17,850 USD (300 hours)
Insurance: 5,000 - 10,000 USD
Hangar: 1,800 - 12,000 USD / year

(source)

It's estimated to cost you 29,650 USD per year to maintain your new Cessna 172.

Your numbers all look pretty high.  Where I am, hangars are 500 USD/month, 6K/year. Insurance varies with pilot experience and hull value. I have 900 hours and instrument rating. My plane is not valued at 200K, but insurance is 1000/yr.   2000/yr might be reasonable for a higher value plane.  Maintenance - 100 hours of fliying is a lot for many pilots, say 75/hr with fuel, landing fees, oil, that's 7500. Annual maintenance another 2000.  If one had the plane for business use, then the flying hours would likely go up, along with the other numbers.

@CrossRoads Yes, these numbers can vary a lot. Thanks for the comment though. I got them from an Honeywell Aerospace infographic (https://aerospace.honeywell.com/~/media/6fddf3a8fea64445b5e79e629f255a09.ashx)

I think your maintenance costs are way too high. I have the same plane as Crossroads and we pay $4k/yr for insurance (it's a club with 9 members). Annual is $1k/yr on a good year.. Unless you fly in Europe or to big airports you don't pay landing fees, so your maintenance costs are off by a good bit. Realistically your costs for a new-ish 172 would be about $12-15k/year, not $30k.

"Airplanes are times more complex than cars." Looks like you're missing a word here, probably "many".

Simple: short production numbers. If light airplanes were as common as motor cars, manufactured by the millions, they would be much cheaper...

New planes?  They are indeed luxurious, with panels that make a modern car look simple. Have you seen the multipanel displays that are going in? The instrument panel alone is worth more than a new car.  A car can be assembled on an assembly line in about a day, yes? With components all trucked in from here & there feeding an assembly line. Planes don't sell in that kind of quantity.
Look at the data here:
http://www.fi-aeroweb.com/General-Aviation.html

As of 2019, there are more than 446,000 general aviation aircraft in the worldwide fleet, ranging from small training aircraft and helicopters to intercontinental business jets. About 213,000 of these, or 48%, are based in the United States. There are more than 5,000 public airports in the U.S. vs. fewer than 400 airports served by commercial airlines.
2019 Worldwide plane deliveries: 2,443.

Can manufacuturers dwarf that number.

Toyota’s motor vehicle production reached almost nine million units in the fiscal year ended March 31, 2019.
  https://www.statista.com/statistics/267272/worldwide-vehicle-production-of-toyota/

And that's just 1 manufacturer. VW, GM, Ford, they're all huge.
Another factor: Just about anyone can get a drivers license and drive a car, with many multi-car households. It takes a fair bit of dedication to get a pilots license, months of training, and then ongoing practice and a minimum of bi-annual recurrency testing/training to keep that license, along with keeping somewhat medically fit. And then supplies like charts that are updated every 6 months, down to 2 weeks for approach plates.  When's the last time you updated a car map?  I only use the GPS moving map in my car.
Older planes can be retrofitted with the latest and greatest avionics, with equipment costing more than a new car (example, a new Avidyne IFD550 GPS/NAV/COM can be over $15K USD - plus installation).
http://www.nexairstore.com/new-avionics/gps-nav-com/
An engine flown for 100 hours a year can last 20 years, and then it can be overhauled, or replaced for $50K or more.
http://www.airpowerinc.com/productcart/pc/engines.asp?searchParm=ALL&catID=33
Planes get inspected annually, and much more thoroughly than the inspection a state may impose on car owners for 39 a year. Some states there is no inspection at all.
Aluminum planes can be repainted, a thorough process where by everything that can be removed is, and all the bits cleaned & repainted individually, then reassembled. A paint job can cost the same as buying a small car.  Repainting a car by comparison is dirt cheap and is done in a few days,
https://www.maaco.com/services/auto-painting/
Vs 6-8 weeks for an airplane for disassembly, prep for stripping, cleanup where stripping missing, alodyne to treat the aluminim, primer, finally paint, and then reassembly.  Composite planes, I don't know the process.
Operating: planes use a lot more fuel per hour to cover ground a lot faster.
Say you had a 180 HP plane burning 10 gal/hr and going 125 mph ground speed (into some headwind).
To go 300 miles will take about 2.4 hours, 24 gallons of gas at roughly 5/gallon = 120.
A car might average 60 miles/hr, taking 5 hours, and getting 30 mpg, so 300 miles = 10 gallons @ 3/gallon, so 30 in gas.
That 300 mile road distance is almost always fewer air miles.
Example, it is 183 nautical miles (210 statute miles) to fly from my home airport to Bangor, Maine, a trip I am making in the next month or two.
1 hour, 26 minute flight. ~15 gallons of fuel $75.
(www.skyvector.com)
Driving, 245 miles. 3 hour, 44 minute flight. $25 in fuel plus tolls.
(maps.google.com)
Longer if there are construction delays (and it is summer construction season where I am) or traffic delays from accidents.
The time savings: 2 hours and 18 minutes. What's your time worth? $50/hr? So $115 saved by flying. Makes up for the difference in fuel cost.
Am I taking my 1973 plane with new paint job and avionics? Or my 2016 car with moving map?  The plane for sure.


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If I were to build a J3 cub twice the size of the original using the same CG would it fly?

How big of an engine would I need, if the original is 65hp?
What would be the weight of the up-scaled plane? Simply twice the
weight?
What would be the stall speed?
How fast would it go?  75mph?
Is there a software program to test aircraft design?

This isn't exactly a duplicate, but there's a lot of relevant information: What are the physical laws for upscaling an RC model to 1:1?

3 good answers from 3 smart people give 3 different HP ratings: 320, 520 (minimum) and 735. There's a lot of math and physics involved and simple "back of the envelope" scribbling won't get you accurate answers!

@Freeman Very right you are. Of course it is possible to construct larger aeroplanes than the J3, we see them flying around very regularly. Point is that linearly amplifying all dimensions does not work, the larger plane needs to be re-engineered with all the design choices and computations that come with that.

@Freeman remember power requirement hinges on weight and drag.  The weight factor is more difficult to predict without detailed analysis, as it will hinge on strength requirements AND additional fuel/cargo/pax. A faithfully scaled (2x) J3 will need to go around 40% faster. I would go with 8x power as an off the cuff. The 74 foot wingspan Ford Trimotor weighs in a good reference, sporting 3x300 hp to lift around 10,000 lbs and cruise at around 100 mph.  The design matches the J3 very well. (Late 1920s).

Unfortunately a simple scale-up would fall victim to the square-cube law: all dimensions times two means the volume and weight are 2x2x2 = 8 times the original, while the wing area is 2x2 = 4 times the original. It is the wing area that keeps the aeroplane up. It is the reason why birds cannot get any larger than they already are.
Someone in the 1920s analysed this and predicted that the maximum size for aeroplanes would be about that of the DC-3, but improved technology has proved that prediction wrong. A B777 has a way higher wing loading than a DC-3 of course. The square-cube law is not a law, it can be defeated by engineering. Point is that linearly amplifying all dimensions does not work.

How big of an engine would I need, if the original is 65hp?

Minimal: 8 x 65hp = 520 hp. We need a serious amount of extra thrust.
From this answer:
$$D =  C_{D0} \cdot \frac{1}{2}  \rho  V^2 \cdot S + \frac{2  W^2}{\pi  A e \cdot  \rho  V^2 \cdot S} \cdot $$

$C_{D0}$ would stay the same if everything is scaled up
S = wing area = 4 times as high
W is 8 times as high.

$D_2 = D_1 \cdot (4 + 8^2/4) = D_1 \cdot 20$
So you need 20 times as much thrust at the same speed.
Increase V and the first factor increases, second factor decreases. There are of course many optimisations possible (weight saving, wing loading increase etc), but that goes beyond the scope of the question.

What would be the weight of the up-scaled plane? Simply twice the weight?

Eight times that of the original.

What would be the stall speed?

Higher than that of the original because the wing loading is higher. Depends on the flaps & slats installed.

How fast would it go? 75mph?

Yeah it might..but it would probably need to go faster as @xxavier mentions.

Is there a software program to test aircraft design?

FlightGear is an open source flight simulator that allows you to link your own flight dynamics model into it.

Comments are not for extended discussion; this conversation has been moved to chat.

In general, the power required by an airplane is proportional to its weight times its speed. The weight scales with the cube of the linear dimension, so the bigger J3 would weigh $ 2^3 = 8 $ times more. Since the airspeed is squared in the formula for lift, the exponent for the proportionality of lift should be 1/2.
Now you have two powers for the scaling, 3 for the weight (or mass, in this case) and 1/2 for the speed. As the weight and speed are multiplied in order to get the proportionality of the power required, I add the exponents, and the result is that the bigger J3 plane would need an engine of $ 2^{3.5} = 11.3 \times 65 = 735 $ hp.
Concerning the stall speed, and as airspeed scales with a power of 1/2, and the scale factor is 2, if the 'small' J3 has a stall speed of –say– 50 mph, you'll have a stall speed of $ 50 \times \sqrt{2} = 71 $ mph for the 'big one'... The cruise and maximum speeds will also scale with $ \sqrt{2} $...
Of course, all these are rough estimations... To start with, the weight does not grow exactly with the 3rd power of the linear dimension...

Yes, it would fly.
Scaling up a successful design has worked well in the past. Just two examples:

When Howard Hughes wanted a fast airliner for his TWA airline, he turned to Lockheed and asked them to design one on the basis of the wing of their P-38 fighter. The result was the Lockheed Constellation (the picture below shows the military version C-69).

In the summer of 1940, Willy Messerschmitt was asked to design a cargo glider with the payload bay the size of a Reichsbahn cargo car. He used the largest aircraft he had designed so far, the M-20, and scaled this up. The result was the Me-321, which was a considerably better design than the analog Junkers design Ju-322, which was designed from scratch and turned out to be uncontrollable.

What would be the weight of the up-scaled plane? Simply twice the weight?

No, it will be much heavier. The structure will have to be beefed up for the same reason that an elephant needs big, sturdy legs while an ant can use spindly legs and still carry a multiple of its own weight. Normally, this factor is the length factor by the power of 2.3 or 2.4. Since the aircraft will be hollow on the inside, simple cube laws which apply for solids will give too high a value.
If you look at existing aircraft, you will find that wing loading increases with size. That makes sense: The wing of an aircraft twice as big as the original will have four times its area. Its mass, however, will be five times bigger according to the exponential weight growth.
Engine size should be selected so the power to weight ratio is maintained. Your twice-as-big Piper J-3 will need 320 hp to be useable.
Since it is bigger and has a higher wing loading, the upscaled J-3 will also fly a bit faster. Stall speed will be 10 - 12% higher due to the higher wing loading. Friction drag will be smaller relative to its wetted area since it flies at a higher Reynolds number. The higher wing loading will again result in higher maximum flight speed, if only because the optimum point of lowest drag is at a higher dynamic pressure. But don't expect too much: I would expect that the top speed is just 8% higher. Maybe 10% if you are lucky.

Is there a software program to test aircraft design?

Yes. Many. But no freeware.

The Constellation does not look like a linear upscale of the P38...of course upscaling is possible, with many technological changes and optimisations.

@Koyovis Look again.  Look at the shape of the wing and stabilizer, the shape, size and position of the outboard fins and rudders.  The center fin and rudder were probably added to account for using four engines (and the possibility of a failure much further outboard) instead of two.

@Koyovis: Only the wing was used for upscaling (obviously).

@ZeissIkon I looked again, and counted the number of fuselages per aircraft.

@Koyovis I see one each.  The P-38 tail booms weren't transferred over because the fuselage of the Connie was longer for cargo/passenger capacity.

@ZeissIkon Ah really? So they were different?

Back in 1973, Air Progress magazine published an article on a proposed three times scaled-up Piper J-3 Cub with 500 hp, original top speed, 200 mile range, etc. You can find the article at:
http://sbiii.com/cyclops/stmott-3.html#kngkngcb

Nice find. The author realises that construction methods must change, and that the MTO must be drastically limited, which is the main challenge.
*... so we've decided that the wing construction should be a little different from the normal J-3's.  By using plywood to cover the wing...
Since we don't want to lose that good old Cub fee1, the wing loading must remain just as it is on the normal airplane, around 6.8 pounds per square foot.  This gives us a maximum permissible gross weight of 10,924.2 pounds.*

Wow! Thankyou for all the reply's, I have learned a lot so far. I am going to have to learn some more math I can see. there is enough info here to get me started Thankyou! I am going to have some more question later, after I do some math caculations.

@Dreamer It is of course possible to construct an aeroplane twice the size of the J3 and that looks like the J3 from the outside. And the weight won't be 8 times to start with, that is only the case if all dimensions are scaled up with factor 2.

@Koyovis - your bit of the quote about keeping the wing loading at about the same weight/area ratio is the key to success if this were a real project.  Modern materials, etc. - assuming of course there is a (super extra large) budget for it

The Piper Cub J3 is an excellent place to start in aircraft design.  It represents all that was learned in the first 40 years of aircraft design as a tractor driven high wing mono plane with a properly proportioned and placed vertical and horizontal control surfaces. It is no accident this was the first trainer aircraft for many aviators.
Undoubtedly this design can be scaled up, but beyond the actual build it may be beneficial to look for the J3's attributes in other larger aircraft.  The high wing gives improved stability in pitch and roll.  This is a plane that wants to stay "upright".  Now look at the 132 foot wingspan C130 Hercules cargo plane.  Yes, a lot of Piper Cub there.  And that large tail keeps it steady and can handle a wide range of CG.
So, for your scale up, plan on needing more power, and high lift devices such as slats and flaps to keep landing speed at a minimum.  Also, evaluate strengthening requirements for fuselage and especially the wing to bear the heavier loads. And check out existing designs in the size range you desire.
And if you are really motivated, here is another Hercules, the Hughes H-4, aka "Spruce Goose".  At 320 feet, it's wingspan is more than 9 times that of the J3.  See if you can pick out the similarities.


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Which are more efficient in putting out wildfires: planes or helicopters?
A lake is near where both helicopter and plane can access the water and fire.
Which aircraft (if you had access to only one) is more efficient in putting out wildfires? Which aircraft is preferred to fly for this? What cost less to operate and which aircraft can deliver the most water per load and on the average?
I have added this link below on how this airplane delivers water. Sorry it is longer than 4 minutes. In the video at 4:49 the plane can touch down and refill without stopping.

If one type was better, the other type wouldn't exist.

Please, who makes that airplane and where was the video shot.  They seem to be on to a very good design!  (A smaller prop amphibian also is seen).

@RobertDiGiovanni https://en.wikipedia.org/wiki/Beriev_Be-200

@DavidRicherby - not true at all. Both have subtly different uses, as shown in answers. So one is better sometimes, the other at other times.

@Tim You're completely agreeing with me...

@DavidRicherby - it's the way you tell 'em! Bit like chalk and cheese.

Note that efficiency in fire-fighting is not always the driving factor: Air tanker drops in wildfires are often just for show

How far from the fire is the nearest helicopterable water source?  How far is the nearest fixed-wing-scoopable lake?   What's the particular kind of fire?  How far is the nearest fixed attack base capable of mixing and handling fire retardent?  Do you have access to large waterbombers like DC-10 or 747?

Both have their respective uses. A fixed-wing carries far more water (3,200 gallons or 12 tons for the Be-200 vs 100-700 gallons for helicopters), but takes more local infrastructure to operate, and is more difficult to discharge exactly where you want it. A helicopter is more precise, but carries less water. A heli is quicker to load up and discharge near the water, but a fixed-wing flies faster, so is better suited to delivering water or a custom mix a longer distance away.
You want both, not just one. You'd use helicopters to suppress smaller fires if possible before they get out of control, attack larger ones at specific spots, and fixed-wings to lay down longer lines of fire retardant against developed wildfires. Larger helicopters also support fixed-wings in this. Which is "better" if you could only have one kind... that's not so much an aviation question as a fire-fighting one.
As for cost, helicopters are normally less cost-effective, being far more expensive to operate per ton-mile. A large aircraft is more expensive in absolute terms. Being smaller and less demanding of the infrastructure, helicopters can be stationed in a more critical location and react more quickly.
It's not a worse/better question, it's tools for the job. Fixed-wing fire-fighting aircraft are more specialized, with their internal water storage, while fire helicopters can be as basic as a light utility heli carrying a water bucket.

During a local (San Jose, CA) fire in hilly terrain I observed the following division of labor: an airplane laid down strips of pink fire retardant. A chain of three or four helicopters ferried water from a nearby reservoir or lake, with only a few minutes spacing between helicopter arrivals at the fire.

Another example that I've seen: a helicopter was able to easily get water from a nearby small lake surrounded by hills that'd make it difficult to get a Piper Cub to it.  And that helicopter could have been used for e.g. search & rescue just by unhooking the water bucket, unlike the special-purpose fixed-wing aircraft.

question revised

Helicopters may cost more per ton-mile, but usually require far fewer miles to do the job.  I watched a variety of aircraft knock down an urban-interface fire: in an hour, the helicopter made twelve runs of a mile each pulling water out of a nearby stream, a gaggle of floatplanes made three ten-mile runs to a lake, and the big tankers made just a single run each, from an airport 30 miles away.

@Mark With a body of water nearby, helicopters win, but you don't always get that luxury. Note that purpose-built fire aircraft like DHC-515 or Be-200 can scoop water bodies as well.

Therac is right but pools for helicopters are built for this purpose. Additionally, helicopters can grab water from swimming pools, that are very common in some places.

It depends on what you are trying to do.
The planes aren't doing direct attacks.  In other words, they don't usually drop on fires.  They lay down lines of fire retardant or water as a fire break.  The fire burns up to the line, slows or stops, and ground crews can go in and stamp out hot spots.  The big Erickson helis can also carry retardant or water and perform a similar function.  Since helis can be more precise, they can hit specific hot spots and cool off the fire.
Direct attacks can be used in some cases, but it's not really a great way to extinguish a fire.  Actually putting a fire out isn't going to happen with a plane or a heli.  They may succeed in cooling the blaze but they probably won't stop it.
Simple version: the fire goes out when it runs out of fuel.  It runs out of fuel when there is nothing left to burn.  A bomber doesn't carry enough slurry or water to snuff out a fire that is worthy of national news coverage.  The strategy is to contain the blaze with retardant or water and allow it to burn up to lines and let crews and fuel exhaustion take care of it.


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What kind of Chinook helicopter/airplane hybrid is this?
I had this question: Could this helicopter's body be a flying wing? and this picture came up.
What kind of helicopter is this?
 What is the extra wing for?

That's just Boeing trying to decide whether to build a plane or a chopper next.

@Mast The project was abandoned because the marketing department just couldn't do anything with "plopper".

@DavidRicherby ... thanks now I have Arnie yelling "Get to the plopper!" stuck in my head.

It‘s a BV-347 technology demonstrator:

In 1969, the Army and then-Boeing Vertol entered into a partnership to improve upon their cargo helicopter fleet, and thus the BV-347 was borne from a CH-47A Chinook that was provided by the Army as a technology demonstrator, according to retired CW5 Jim Kale, who now serves as a tour guide for the U.S. Army Aviation Museum.

Primary objectives as found in above article:

The BV-347 was tested for more than two years and flew about 350 hours with the intention to meet five primary goals: improve flying qualities with external loads and instrument flight; im-prove stability, control and maneuverability in hover flight and forward flight with high gross weight; reduce rotor noise; reduce vibration stress; and create a modern cockpit environment with improved instrument and navigation displays.

The wing was intended to provide extra lift (works in forward flight only, of course), but apparently didn’t add net benefit:

The wings were added as a means to improve lift, and although the addition served its purpose, the added weight canceled out any positive effects it provided [...].


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When are more engines better than not on a plane?
A model of this airplane was used to carry a space shuttle. In this picture below the airplane had a few extra engines added since then.
How is the number of engines decided on an airplane? Why not use bigger engines?

I think this picture maybe fake? ANTONOV 925

When are more engines better than not on a plane? When one or more of them has quit working.


text:
Calculate speed by turn angle and distance
I'm trying to combine two rate of turn formulas, which I found it online this and this.
The first is ROT = 1091 * tan(bank angle) / speed in knots
The second is ROT = speed in knots / radius
By combination,
1091 * tan(bank angle) / speed in knots = speed in knots / radius
speed in knots = [1091 * tan(bank angle) * radius] ^ 0.5

By finding radius, use the simple trigonometry formula c^2 = a^2 + b^2 - 2ab * cos(C)
distance^2 = radius^2 + radius^2 - 2 * radius * radius * cos(bank angle)
distance^2 = 2 * radius^2 * (1 - cos(bank angle))
radius = distance / [2 * (1 - cos(bank angle))] ^ 0.5

Therefore
speed in knots = [1091 * tan(bank angle) * [distance / [2 * (1 - cos(bank angle))] ^ 0.5]] ^ 0.5

Below is the calculation I got. It shows higher bank angle, higher speed and lower turn radius. Does this make sense if the angle-of-attack is fixed? Or I have some wrong calculation? Thanks.
bank angle:     5      10     20     35
distance_in_NM: 2.33   2.33   2.33   2.33
turn_radius:    26.67  13.35  6.70   3.87
speed_in_knots: 50.45  50.67  51.58  54.36

-------------EDIT-------------
The first is still the same. ROT (°/sec) = 1091 * tan(bank angle) / speed in knots
The second is ROT (°/min) = 0.955 * speed in knots / radius. By converting it to °/sec and thanks by Gerry suggestion, ROT (°/sec) = [0.955 * (speed in knots / 60)] / radius
To combine these two formulas,
1091 * tan(bank angle) / speed in knots = [0.955 * (speed in knots / 60)] / radius
speed in knots = [1091 * tan(bank angle) * radius * 60 / 0.955] ^ 0.5

As a result,
speed in knots = [(1091 * tan(bank angle) * 60 / 0.955) * [distance / [2 * (1 - cos(bank angle))] ^ 0.5]] ^ 0.5

Another attempts with different bank angles. Still higher bank angle, higher speed and lower turn radius.
bank angle:     5       10      20      35
distance_in_NM: 2.33    2.33    2.33    2.33
turn_radius:    26.67   13.35   6.70    3.87
speed_in_knots: 399.91  401.64  408.81  430.89

A 3-mile turn radius at 50 knots & 35 degrees of bank looks too large, as does a 26 mile (!!!) turn radius, even at just 5 degrees of bank. I suspect there is a units conversion gone wrong someplace.

@Ralph J are the two formulas correct?

The first equation gives ROT in deg/sec. The second gives ROT in deg/min. There's a factor of 60  that needs to be addressed.

As described the second equation is an approximation.  Consider 60 kts and a radius of 1 NM.  In one minute at 60 kts, you would travel 1 NM (along the curve).  That would mean an angular change in direction of 1 radian. Since 1 radian is 57.3 degrees. that would less than the 60 deg per minute provided by the equation. More correct would be ROT_in_deg/min=0.955 * Speed_in_Kts / radius_in_NM.  You need to devide that by 60 to deg/sec.

You should edit your values to properly reflect significant figures Your numbers imply that you are capable of measuring distances with accuracies in the tens of picometers, or millionths of an inch... (relevant xkcd)

VTC based on the problems with the question as described in quiet flyer's answer (and @randomhead's comment).

@Gerry Thanks. I adjusted the answer.

@randomhead You are right cuz I calculate the number by coding. Thanks for your reminding. I adjusted the numbers

The pic is not very clear. Is it a top view of an aeroplane flying a coordinated turn? Then, which distance do you want to compute at which specific point in time? How is aeroplane bank angle now an angle determining 2 points on a circle?

If this question gets any more editing then that would be a good time to add the "flight-dynamics" tag

The issue here is the mixing of units without understanding the need to use all the same units, and ensure that they cancel out. To do that unit conversion constants (like 1 nm = 6080 feet, etc. etc.) must be included. See my answer below.

Below is the calculation I got. It shows higher bank angle, higher
speed and lower turn radius. Does this make sense?

If we specified that angle-of-attack were fixed, so that airspeed increased as a function of wing loading, then it would make perfect sense that banking steeper would result in both a higher airspeed, and a smaller turn radius.
But that doesn't appear to be what you are doing here.  You don't have any formula relating to wing loading.  You have too many outputs and not enough inputs -- the only input appears to be bank angle.  That's not enough information to draw any valid conclusions about airspeed or turn rate.
If turn rate and bank angle were both inputs, then we could compute the airspeed.  But that doesn't appear to be what you are doing.
It appears that you need to start over with a fresh "clean sheet" of paper and fundamentally re-think your approach here.

Thanks for the answer. Yea. Assumption for the angle-of-attack is fixed here as I use these formula to make some simulations.

@PakHoCheung -- I suggest editing the question to include that information--

@PakHoCheung -- I suspect your analysis may be entirely correct, I haven't gone through it in enough detail to be sure-- you just didn't include all the needed information, in the actual question--

Edited to include that information. Thanks:)

If we remain in the SI metric unit system, this answer finds 2 equations:

$\omega = \frac{g}{V} \cdot \tan (\Phi)$
$ \omega = L \cdot \sin(\Phi) / (m \cdot V)$

With $\omega$ in rad/sec.
Equation 1. is for an aeroplane making a coordinated turn, keeping altitude. Equation 2 is not concerned with keeping altitude. Your second equation ROT = speed in knots / radius boils down to V = $\omega \cdot$r, take any V and any r.
So $V/r = g/V \cdot \tan (\Phi)$ finds $$ r = \frac{V^2}{g\cdot \tan (\Phi)} $$.
No need to compute a distance, however a protection against divide-by-zero is required.

First understand the physics.
Rate of turn (ROT) is the number of degrees of heading change that an aircraft makes in a set period of time. It is measured in degrees per second or degrees per minute. The formula for ROT is simple. It is the radial G divided by the aircraft true air speed.
Radial G is the acceleration in the direction you are turning. It is measured in acceleration units, distance per time squared (say meters per second squared), and Velocity of course is in Distance per time (Meters per second). When you perform this division the meters and one of the seconds units cancel each other, and what you have left is something per second. By definition, this is an angle measured in radians ( approximately 57 degrees )
The first formula you are using
1091 * tan(bank angle) / speed in knots
is derived from the above by assuming level flight (no climb or descent) where the radial G is the horizontal component of the acceleration due to the lift from the wing at a specific bank angle (radial G = tan (Bank angle) * 9.81 m/sec2), and  converting knots (nautical miles/hr) into meters/sec.
So One nautical mile is 6080ft and one meter = 3.28 feet so nautical mile is
6080 / 3.28 = 1853.65 meters
But a knot is nautical miles per hour, so to get m/sec we also need to divide by the number of seconds in an hour (3600)
(1 knot = 1853.65 meters/hour / 3600 sec/hour), and then, since the answer would be in dimensionless radians/sec, and we want it in degrees/sec, multiplying by the number of degrees in a radian (57.2958 degree/radian), to get 1091.
9.81 * 3.28/6080 * 57.2958 * 3600 ~= 1091
That's where this formula came from.
The second one is simpler. It is based on the fact that there are 360 degrees in a circle, and that the circumference of a circle is 2 * Pi * radius
So if your airplane travels at velocity V (in knots) it will travel completely around the circle (360 degrees), in amount of time = (Distance / Velocity) hours (because we are specifying V in nautical miles per hour).
The rate of turn is the number of degrees per unit time, so we just need to divide 360 degrees by that amount of time, and then also divide by 3600 to get from degrees per hour to degrees per second...
 ROT = 360 V / (2 * Pi * radius * 3600)
 ROT = V / (20 * Pi * radius):
 ROT = V / (62.8318 * radius):   V in knots, radius in nm

This formula is dependent on entering the Velocity in Knots, and the radius in nautical miles. if you want to enter the radius in feet, then you need to convert that as well (one nm = 6080ft) and the formula becomes
 ROT =  96.7 * V / radius:   V in knots, radius in feet

Remember, all of this is only true for an aircraft in a level (no climb or descent) constant bank angle turn). If you are turning in the vertical, the assumptions in this about the radial G required to maintain level flight at a specified bank angle will no longer be true.

Thanks for the explaination. I did wonder how the 1,091 come from. Super clear now

Units are important!


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How do I find out if my aircraft satisfies a particular ICAO equipment/surveillance/PBN code?
I have a new, G1000-equipped C172S that just arrived, and I’m trying to figure out the specific ICAO equipment/surveillance/PBN/etc. codes for the aircraft.
I’ve been trying to find out where to find this information via Google, but I haven’t found much that tells me where, specifically, to find this information for my specific aircraft.
Can someone please elaborate, in detail (“ELI5”), where I find this information?
For Example: Where do I find out if my aircraft has GNSS, SBAS, APV with SBAS, RNAV 1, RNAV 5, etc.
Thanks.

Welcome to aviation.SE! This question has a generic answer; this one is specifically about the C172 and G1000.

Not authoritative, but for the 2019-2021 model C172S/G1000 planes at my flight school, I was told to use ICAO Equipment B,G,R,S; ICAO Surveillance B2,S; and ICAO PBN B2,C2,D2 in ForeFlight.